Is there any theorem which says anything like $SN_G(D)=G$ where $S$ is a $p$-Sylow subgroup and $D$ is the intersection it has with some other subgroup?
I know the first that will probably spring to mind is the Frattini argument, but the thing is in the proof where I saw this the author is using this to show that $\cap_{g \in G}S^g$ is normal in $G$ (to get a contradiction as we assumed $G$ was simple) but then surely we can't use the Frattini argument as it would require $S$ to be normal.
Let G be a group of order $p^nq$ $n \geq 0$ p and q are primes Then G is not simple.
Edit: To add more clarity here is the full argument of the section I'm stuck on .
Proof: Wlog assume $p \neq q$ Suppose G is not simple
( NOTE: I'm going to skip the parts for when n = 1 and where $n>1$ but has trivial intersection as they are separate cases and their arguments don't play any role in the part of the proof I'm interested in.)
So suppose $n>1$ and pick any two Sylow subgroups $S \neq P$ s.t. $P\cap S=D$ is largest possible.
Then $D<S \Rightarrow D<N_S(D)<N_G(D)$
and also
$D<S \Rightarrow D<N_P(D)<N_G(D)$
It follows that $N_G(D)$ can not be a p group since if so it lies in some P-sylow subgroup T and Then $T\cap P$ $\geq N_G(D)$$\geq N_P(D)>D$ $\Rightarrow T=P$ But then similarly $S\cap P=S\cap T<D \Rightarrow \Leftarrow$
Thus a q sylow subgroup Q lives in $N_G(D)$ Then $SN_G(D)=G$ and so if we pick any g it will be of the form $g=sx,s\in S, x\in N_G(D)$
Then $S^g=S^{sx}=S^x\geq D^x=D$ and thus D lies in in every P sylow subgroup of G $\Rightarrow 1<D \leq \cap_{g\in G} S^g$ is normal in G which is a contradiction.
So, this is how Isaacs proves the theorem you mentioned. Strategy and notation look very similar to the proof in your notes, so this is likely where your teacher took it from.